How to raise -1 to non-integer powers

Question

How do you calculate $(-1)^x$ where $x$ is some real number. For example, what is $(-1)^{\sqrt{5}}$. This question came as I was trying to computer $e^{i\pi a}$ where $a$ is irrational.

Answer 1

I used this very successfully on page 7 of http://zakuski.utsa.edu/~jagy/papers/Intelligencer_1995.pdf

One value is $$ (-1)^x = \cos \pi x + i \sin \pi x $$

That was enough for my article, as this complex number can be used in Gelfond-Schneider. The conclusion is that if $x$ is real, irrational, but algebraic, then $\cos \pi x + i \sin \pi x$ is transcendental. I used one possible version of a contrapositive: I had both (real) $x$ and $\cos \pi x + i \sin \pi x$ algebraic, therefore $x$ was rational.

Note, however, that there are countably infinite logarithms of $(-1),$ so there are countably infinite different values of $(-1)^x.$ That's just life.

Answer 2

In general, $$e^{it}=\cos t+ i\sin t$$ This is known as Euler's formula. If we apply it to your example, we get $$(-1)^{\sqrt{5}}=\cos(\pi\sqrt 5)+i\sin(\pi \sqrt 5)$$ Since $-1$ has other representations as $e$ to some power (we can add any multiple of $2\pi i$ to its exponent), other numbers could also be called $(-1)^{\sqrt{5}}$, so it's better to work with $e^{i\pi\alpha}$.

Answer 3

In general, for complex numbers (with $a \ne 0$) $a^b$ is defined as $e^{b \log(a)}$. However, $\log(a)$ is a multivalued function, and therefore so is $a^b$.